# Laguerre's Differential Equation

• September 18th 2010, 08:51 AM
11rdc11
Laguerre's Differential Equation
I need just a little help with this question. I almost solved it except for a minor detail at the end.

Here is the question.

The equation

$xy''(x) + (1-x)y'(x) +my(x) = 0$,

where m is a nonnegative interger, is called Laguerre's differential equation. Show that for each m, this equation has a polynomial solution of degree m. These polynomials are denoted by $L_{n}(x)$ and are called Laguerre polynomials. The first few Laguerre polynomials are

$L_{0}(x) = 1$
$L_{1}(x) = -x +1$
$L_{2} = x^2-4x+2$

Ok so I did a power series solution for the problem and arrived at

$a_{0} = constant$

$a_{1} = -ma_{0}$

$k \geq 1$ $a_{k+1} = \frac{a_{k}(k-m)}{(k+1)(k) +(k+1)}$

so

$a_{2} = a_{0}\frac{(-m)(1-m)}{4}$

$a_{3}=a_{0}\frac{(-m)(1-m)(2-m)}{(4)(9}$

$a_{4} = a_{0} \frac{(-m)(1-m)(2-m)(3-m)}{(4)(9)(16)}$

and this is where I get confused.

I see that there is a factorial pattern so I'm trying to use factorial notation to condense.

so would it be

$\sum_{n=2}^{\infty} \frac{(-m)_{n}}{(1)^2_{n}}x^n$

or

$\sum_{n=2}^{\infty} \frac{(-m)_{n}}{(1^2)_{n}}x^n$

So would the final solution be

$m \geq 0$

$y(x) = a_{0} -ma_{1}x +$ whatever series is correct?

This solution would show depending what m is chosen that the answer would cancel out certain terms.
• September 18th 2010, 08:59 AM
11rdc11
The second part of the question is this

Use the results to obtain the 1st few terms in a series expansion about x=0 for a genreal solution for x > 0 to Laguerre's differential equation for n = 0 and 1. Does this just mean plug in 0 and 1 into the general solution I found earlier?